Question:

If a body is thrown vertically upwards from the ground with a velocity of 20 m/s, then its displacement during the last second of upward motion is (Acceleration due to gravity \(g=10\,m/s^2\)):

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Displacement in last second = total height - height at (total time - 1) sec.
Updated On: Jun 2, 2025
  • 5 m
  • 10 m
  • 15 m
  • 20 m
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The Correct Option is A

Solution and Explanation

Total time to reach max height: \[ t = \frac{u}{g} = \frac{20}{10} = 2\,s \] Displacement in last second of upward motion (from \(t=1\) to \(t=2\)): Displacement from ground at \(t=1\): \[ s_1 = ut - \frac{1}{2}gt^2 = 20 \times 1 - \frac{1}{2} \times 10 \times 1^2 = 20 - 5 = 15\,m \] Displacement at \(t=2\) (max height): \[ s_2 = 20 \times 2 - \frac{1}{2} \times 10 \times 4 = 40 - 20 = 20\,m \] Displacement during last second: \[ s_2 - s_1 = 20 - 15 = 5\,m \]
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